The mean serum cholesterol level in the population of males who do not develop heart disease is known to be 219 mg/100 ml. Suppose you perform a study to investigate whether the mean serum cholesterol levels in males who do develop heart disease is higher than that of the disease-free population. In a sample of 28 men who have heart disease, the sample mean serum cholesterol level is 232 mg/100 ml with a sample standard deviation of 38 mg/100 ml. Perform the appropriate test at the 0.10 level of significance.
a. Ho :
b. Ha :
c. Critical value(s) =
d. Test statistic =
e. Approximate P-value =
f. Decision:
Interpret this decision:
a non-normal population is determined to have a mean of 60 and a standard deviation of 4. Ninety-six percent of all observed values will occur in what range?
Assume a normal distribution.
Ho: mean = 219 mg/100 ml
Ha: mean ≠ 219 mg/100 ml
c. Z > ± 1.645
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion
First, if you have a question, it is much better to put it in as a separate post in <Post a New Question> rather than attaching it to a previous question, where it is more likely to be overlooked.
If it is not normal, the assumptions applied to a normal distribution will be in error. However, for a normal distribution, this is what you would want.
Mean ± 2.055 SD
a. Ho (null hypothesis): The mean serum cholesterol level in males who develop heart disease is not higher than that of the disease-free population.
b. Ha (alternative hypothesis): The mean serum cholesterol level in males who develop heart disease is higher than that of the disease-free population.
c. Critical value(s): To perform the test at the 0.10 level of significance, we need to find the critical value of the test statistic. Since the alternative hypothesis is one-sided (testing for a higher mean), we will use a one-tailed test.
To find the critical value, we need to consult the t-distribution table with n-1 degrees of freedom (where n is the sample size). The sample size is 28, so the degrees of freedom is 28 - 1 = 27.
Looking up the critical value for a one-tailed test at the 0.10 level of significance and 27 degrees of freedom, we find the critical value to be approximately 1.703.
d. Test statistic: The test statistic for this scenario is the t-statistic.
The formula for calculating the t-statistic is:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Plugging in the values from the problem:
t = (232 - 219) / (38 / sqrt(28))
Calculating this expression gives us the value of the test statistic.
e. Approximate P-value: To calculate the approximate P-value, we can use the t-distribution table or a statistical software.
Using the t-distribution table, we find the P-value associated with the test statistic obtained in step d. The P-value corresponds to the probability of obtaining a test statistic as extreme as the one observed (or more extreme) under the assumption that the null hypothesis is true.
f. Decision: If the P-value obtained in step e is less than or equal to the significance level of 0.10, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Interpretation: If the decision is to reject the null hypothesis, it means that there is sufficient evidence to suggest that the mean serum cholesterol level in males who develop heart disease is higher than that of the disease-free population. If the decision is to fail to reject the null hypothesis, it means that there is not enough evidence to conclude that the mean serum cholesterol level in males who develop heart disease is higher than that of the disease-free population.